3.8.72 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\) [772]

Optimal. Leaf size=180 \[ -\frac {2 \sqrt [4]{-1} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-2*(-1)^(1/4)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)
^(1/2)/d/a^(1/2)+(-1/2+1/2*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2
)*tan(d*x+c)^(1/2)/d/a^(1/2)-1/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {4326, 3639, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {2 \sqrt [4]{-1} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(-2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*S
qrt[Tan[c + d*x]])/(Sqrt[a]*d) - ((1/2 - I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c
+ d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[a]*d) - 1/(d*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x
]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*a*(b/f), Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3682

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

\begin {align*} \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\left (-\frac {a}{2}+i a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^2}\\ &=-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^2}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (i a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {2 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.91, size = 210, normalized size = 1.17 \begin {gather*} \frac {i e^{-2 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-1+e^{2 i (c+d x)}+e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\cot (c+d x)}}{\sqrt {2} a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(I*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-1 + E^((2*I)*(c + d*x)) + E^(I*(c + d*x))*Sqrt[-1
+ E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] - 2*Sqrt[2]*E^(I*(c + d*x))*Sq
rt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d
*x]])/(Sqrt[2]*a*d*E^((2*I)*(c + d*x)))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (141 ) = 282\).
time = 48.80, size = 621, normalized size = 3.45

method	result	size
default	\(\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (-i \cos \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right )-i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-i\right )+i \sin \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )+i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right )+i \cos \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-i\right )-i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right )+\cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-i \cos \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+i\right )+i \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+i \cos \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right )+i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+i\right )-\sin \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right )+\sin \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+i\right )+\sin \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right )-\sin \left (d x +c \right ) \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-i\right )-\sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-\sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2} \left (\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, a}\)	\(621\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(1/2+1/2*I)/d*(-I*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-
I)+I*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+I*ln(((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)+1)+I*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)-I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)-1)+cos(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-I*cos(d*x+c)*ln(((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)+I)+I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I*cos(d*x+c)*ln(((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)-1)+I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)-sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))
^(1/2)-1)+sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)+sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
+1)-sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)-sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2^(1/2)*
arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)))*cos(d*x+c)^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(
d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c))/sin(d*x+c)^2/(cos(d*x+c)/sin(d*x+c))^(3/2)/((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)/a

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c)^(3/2)), x)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (134) = 268\).
time = 1.07, size = 620, normalized size = 3.44 \begin {gather*} \frac {{\left (a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + a d \sqrt {-\frac {4 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-16 \, {\left (\sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {4 i}{a d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt {-\frac {4 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (16 \, {\left (\sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {4 i}{a d^{2}}} - 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/4*(a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c)*log(-2*(sqrt(2)*(I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt(a/(e^(2*I
*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-2*I/(a*d^2)) - 2*I*a*e^(
I*d*x + I*c))*e^(-I*d*x - I*c)) - a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c)*log(-2*(sqrt(2)*(-I*a*d*e^(2*I*d*x +
2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*
sqrt(-2*I/(a*d^2)) - 2*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + a*d*sqrt(-4*I/(a*d^2))*e^(I*d*x + I*c)*log(-16
*(sqrt(2)*(a^2*d*e^(3*I*d*x + 3*I*c) - a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-4*I/(a*d^2)) + 3*a^2*e^(2*I*d*x + 2*I*c) - a^2)*e^(-2*I*d*
x - 2*I*c)) - a*d*sqrt(-4*I/(a*d^2))*e^(I*d*x + I*c)*log(16*(sqrt(2)*(a^2*d*e^(3*I*d*x + 3*I*c) - a^2*d*e^(I*d
*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(
-4*I/(a*d^2)) - 3*a^2*e^(2*I*d*x + 2*I*c) + a^2)*e^(-2*I*d*x - 2*I*c)) - 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-I*e^(2*I*d*x + 2*I*c) + I))*e^(-I*d*x - I
*c)/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(1/(sqrt(I*a*(tan(c + d*x) - I))*cot(c + d*x)**(3/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

int(1/(cot(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2)), x)

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